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nekomata comments on Problematic Problems for TDT - Less Wrong
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I don't understand the special role of box 1 in Problem 2. It seems to me that if Omega just makes different choices for the box in which to put the money, all decision theories will say "pick one at random" and will be equal.
In fact, the only reason I can see why Omega picks box 1 seems to be that the "pick at random" process of your TDT is exactly "pick the first one". Just replace it with something dependant on its internal clock (or any parameter not known at the time when Omega asks its question) and the problem disappears.
Omega's choice of box depends on its assessment of the simulated agent's choosing probabilities. The tie-breaking rule (if there are several boxes with equal lowest choosing probability, then select the one with the lowest label) is to an extent arbitrary, but it is important that there is some deterministic tie-breaking rule.
I also agree this is entirely a maths problem for Omega or for anyone whose decisions aren't entangled with the problem (with a proof that Box 1 will contain the $1 million). The difficulty is that a TDT agent can't treat it as a straight maths problem which is unlinked to its own decisions.
Why is it important that there is a deterministic breaking rule ? When you would like random numbers, isn't it always better to have a distribution as close as random as possible, even if it is pseudo-random ?
That question is perhaps stupid, I have the impression that I am missing something important...
Remember it is Omega implementing the tie-breaker rule, since it defines the problem.
The consequence of the tie-breaker is that the choosing agent knows that Omega's box-choice was a simple deterministic function of a mathematical calculation (or a proof). So the agent's uncertainty about which box contains the money is pure logical uncertainty.
Whoops... I can't believe I missed that. You are obviously right.